[PD] Make timer / clock

[sara kolster]

Hey Frank, Matthew and Phillip,

Thanks so much for the input. I succeeded in making a countdown 
combining the expr. and the mod / div objects. This expr is quite fun to 
experiment with ;)

Does anyone know which logical math element I can use  for the following 
problem? I was wondering if there is such a element instead of using 
select or expr.

The object should only give a 0/1 when two separate floats [connected to 
right & left inlet] inlet are 0?

I tried [&&] and [== 0] but these don't act as I want to.
Now I use [expr if($f1 + $f2 == 0, 0, 1] but I was wondering if one 
there is another way to do it.

s


Frank Barknecht wrote:

>Hallo,
>Frank Barknecht hat gesagt: // Frank Barknecht wrote:
>
>  
>
>>But you are also using an unusual way to count. This is hard to
>>explain in ascii-patching, so maybe you want to take a look at
>>attached explanation in Pd-format.
>>    
>>
>
>Uhm, some errors in that patch: Replace all occurences of the word
>"left" with "right". :(
>
>Attached is the fixed version for archive searchs..
>
>Ciao
>  
>
>------------------------------------------------------------------------
>
>#N canvas 248 25 581 856 10;
>#X obj 59 21 cnv 15 300 30 empty empty Counters_in_Pd 20 12 0 14 -233017
>-66577 0;
>#X obj 174 132 + 1;
>#X obj 172 276 + 1;
>#X obj 130 276 f 0;
>#X obj 272 132 + 1;
>#X obj 236 132 f 0;
>#X obj 114 132 float 0;
>#X text 207 133 ==;
>#X text 311 135 ~=;
>#X obj 380 132 + 1;
>#X obj 344 132 f;
>#X text 66 168 Every bang sent to the float will first output the current
>value of the float to its outlet and to the [+ 1] \, which then adds
>1 to the internally stored value of the float *without* outputting
>it immediatly.;
>#X text 64 226 The new value will be emitted on the next bang.;
>#X floatatom 130 307 5 0 0 0 current_output - -;
>#X msg 130 254 bang;
>#X floatatom 172 308 5 0 0 1 next_output - -;
>#X obj 192 497 + 1;
>#X obj 150 497 f 0;
>#X floatatom 150 528 5 0 0 0 current_output - -;
>#X msg 150 432 bang;
>#X floatatom 192 528 5 0 0 1 next_output - -;
>#X msg 166 460 0;
>#X obj 70 340 cnv 15 300 20 empty empty Resetting_Counters 20 10 0
>12 -233017 -66577 0;
>#X obj 67 588 cnv 15 300 20 empty empty From-To-Counters_(Modulo-n)
>20 10 0 12 -233017 -66577 0;
>#X obj 197 757 + 1;
>#X obj 155 757 f 0;
>#X floatatom 155 817 5 0 0 0 current_output - -;
>#X msg 155 699 bang;
>#X msg 171 726 0;
>#X obj 155 784 mod 4;
>#X text 92 634 If you want to count from 0 to a max-value and then
>start over \, use a "Modulo N" counter. Note \, that the [mod] is inserted
>*between* the float and the [+ 1] addition. To count until N use a
>[mod N-1] object:;
>#X text 65 75 Counting in Pd doesn't require externals \, you can also
>count with the builtin objects. The basic idiom to remember is [f 0]
>x [+ 1]:;
>#X text 212 785 <= count from 0 to 3 (where 3 == 4 - 1);
>#X floatatom 111 774 5 0 0 0 - - -;
>#X text 87 376 To start over \, send the new starting value to the
>right outlet of the float:;
>#X text 193 459 <= RESET using right inlet of float;
>#X connect 1 0 6 1;
>#X connect 2 0 3 1;
>#X connect 2 0 15 0;
>#X connect 3 0 2 0;
>#X connect 3 0 13 0;
>#X connect 4 0 5 1;
>#X connect 5 0 4 0;
>#X connect 6 0 1 0;
>#X connect 9 0 10 1;
>#X connect 10 0 9 0;
>#X connect 14 0 3 0;
>#X connect 16 0 17 1;
>#X connect 16 0 20 0;
>#X connect 17 0 16 0;
>#X connect 17 0 18 0;
>#X connect 19 0 17 0;
>#X connect 21 0 17 1;
>#X connect 24 0 25 1;
>#X connect 25 0 29 0;
>#X connect 25 0 33 0;
>#X connect 27 0 25 0;
>#X connect 28 0 25 1;
>#X connect 29 0 26 0;
>#X connect 29 0 24 0;
>  
>
>------------------------------------------------------------------------
>
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